Các câu hỏi tương tự
A=(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)−1/15 . 4^64
bài 1
(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)-1/15*4^64
Bài 2:Cho x+1/x=10. Tính S=x^5+1/x^5
Tính
S=\(-1^2+2^2-3^2+4^2-...+2016^2\)
A=\(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}.4^{64}\)
Mình đang gấp giúp mình nha!
tính: A=(2+1).(2^2+1).(2^4+1).(2^8+1).(2^16+1)+(2^32+1)-(2^64).
9 tháng 8 2023 lúc 21:50
bài 6:tính nhanh
7)1^2-2^2+3^2-4^2+....-2004^2+2005^2
8) (2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}
Đọc tiếp
bài 6:tính nhanh
7)1\(^2\)-2\(^2\)+3\(^2\)-4\(^2\)+....-2004\(^2\)+2005\(^2\)
8) (2+1)(2\(^2\)+1)(2\(^4\)+1)(2\(^8\)+1)(2\(^{16}\)+1)(2\(^{32}\)+1)-2\(^{64}\)
tính : 1^2-2^2+3^2-4^2+...-2004^2+2005^2
b) (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)-2^64
BT7: Tính
\(3,C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)
\(4,D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(5,E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
Cho M =12.(5^2 + 1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)
Và N=5^64.Tính 2.M_N
Cho A=(4+1)(42+1)(44+1)(48+1)...(432+1)
B=464-1
CMR B=3A